# How do you find square root of 138?

Apr 3, 2018

$\sqrt{138} \approx \frac{39019777}{3321584} \approx 11.74734012447073$

#### Explanation:

The prime factorisation of $138$ is:

$138 = 2 \cdot 3 \cdot 23$

Since this contains no squared terms, the square root cannot be simplified and not being a perfect square, it is irrational.

Note that:

${11}^{2} = 121 < 138 < 144 = {12}^{2}$

So $\sqrt{138}$ is somewhere between $11$ and $12$, closer to $12$.

Let us approximate it as $11 \frac{3}{4} = \frac{47}{4}$.

This is actually a very efficient approximation, since:

${47}^{2} = 2209 = 2208 + 1 = {4}^{2} \cdot 138 + 1$

A much more formal way to find such an efficient initial approximation is to be found at https://socratic.org/s/aPLdnFSE

Next, consider the quadratic with zeros $47 + 4 \sqrt{138}$ and $47 - 4 \sqrt{138}$...

$\left(x - 47 - 4 \sqrt{138}\right) \left(x - 47 + 4 \sqrt{138}\right) = {x}^{2} - 94 x + 1$

From this quadratic we can define an integer sequence recursively, using the rules:

$\left\{\begin{matrix}{a}_{0} = 0 \\ {a}_{1} = 1 \\ {a}_{n + 2} = 94 {a}_{n + 1} - {a}_{n}\end{matrix}\right.$

The first few terms of this sequence are:

$0 , 1 , 94 , 8835 , 830396 , 78048389$

The ratio between consecutive terms converges very rapidly towards $47 + 4 \sqrt{138}$

So we can approximate:

$\sqrt{138} \approx \frac{1}{4} \left(\frac{78048389}{830396} - 47\right)$

$\textcolor{w h i t e}{\sqrt{138}} = \frac{1}{4} \left(\frac{39019777}{830396}\right)$

$\textcolor{w h i t e}{\sqrt{138}} = \frac{39019777}{3321584}$

$\textcolor{w h i t e}{\sqrt{138}} \approx 11.74734012447073$